html - Python爬虫，翻页数据怎么爬，URL不变

Question

网址：http://quote.eastmoney.com/ce...
我想爬所有页的名称数据，（这里只有两页），判断有没有下一页的条件该怎么写呢？
代码：

from selenium import webdriver
driver=webdriver.PhantomJS()

url='http://quote.eastmoney.com/center/list.html#28003684_0_2'
driver.get(url)
usoup = BeautifulSoup(driver.page_source, 'xml')
n=[]
while True:
     t=usoup.find('table',{'id':'fixed'})
     utable=t.find_all('a',{'target':'_blank'})
     for i in range(len(utable)):
          if i % 6 ==1:
             n.append(utable[i].text)
          if #停止条件怎么写:
            break
     driver.find_element_by_xpath(r'//*@id="pagenav"]/a[2]').click()
     usoup = BeautifulSoup(driver.page_source, 'xml')

后面这里就不会写了。。。

Answer 1

You can judge the entries on each page. There are 20 entries on each page. If the current page has less than 20 entries, it means that this page is the last page. You should stop after crawling the current page

Answer 2

By the way, doesn’t this form have a jsonp return interface? Why still climb?

Answer 3

It uses the jsonp interface, just take it.

If you have to crawl it, you can only use a simulation page like selenium + phantomjs to get it.

Answer 4

http://nufm.dfcfw.com/EM_Fina...{rank:[(x)],pages:(pc)}&token=7bc05d0d4c3c22ef9fca8c2a912d779c&jsName=quote_123&_g=0.5385195357178545